Question

Figure shown a small block `A` of mass `m` kept at the left end of `a` plank `B` of mass `M = 2m` and length `l` . The system can side on a horizontal rood. The system is started towards right with the initial velocity `v`. The friction coefficient between the rood and the plank is `1//2` and that between the plank and the block is `1//4` . Find

(a) the time elapsed before the block separates from the plank
(b) displacement of block and plank relative to ground till that moment.

Answer 1

There will be relative motion between block and plank and road. So at each surface limiting friction will act. The direction of friction forces at different surfaces are as shown in figure.

Hence `f_(1) = ((1)/(4)) (mg)`
and `f_(2) = ((1)/(2)) (m + 2m) g = ((3)/(2))mg`
Retardation of `A` is `a_(1) = (f_(1))/(m) = (g)/(4)`
and retardation of `B` is `a_(2) = (f_(2) - f_(1))/(2m) = (5)/(8) g`
Since `a_(2) gt a_(1)`
Relative acceleration of `A` with respect to `B` is
`a_(r) = a_(2) - a_(1) = (3)/(8) g`
Intial velocity of both `A` and `B` is `v`. So there is no relative intial velocity. Hence,
(a) Applying `s = (1)/(2) at^(2)`
or `iota = (1)/(2) a_(r) t^(2) = (3)/(16) g t^(2)`
:. `t = 4sqrt((l)/(3g))`
(b)Displacement of block `s_(A) = u_(A)t - (1)/(2) a_(A) t^(2)`
or `s_(A) = 4u sqrt((l)/(3g)) - (1)/(2) . (g)/(4) . ((16l)/(3g))` `(a_(A) = a_(1) = (g)/(4))`
or `s_(A) = 4u sqrt((l)/(3g)) - (2)/(3) l`
Displacement of plank `s_(B) = u_(B)t - (1)/(2) a_(B) t^(2)`
or `s_(B) = 4u sqrt((l)/(3g)) - (1)/(2) ((5)/(8) g) ((16l)/(3g))` `(a_(B) = a_(2) = (5)/(8) g)`
or `s_(B) = 4u sqrt((l)/(3g)) - (5)/(3) l`